Motion in a Straight Line

To find the average velocity given the motion of a person traveling along a straight line with two different velocities, we start by calculating the average velocity ($ v_{\text{avg}} $) using the total distance traveled divided by the total time taken.

The distances and velocities are given as follows:

Distance $ x $ at velocity $ v_1 = 5 \, \text{m/s} $

Distance $ \frac{3}{2}x $ at velocity $ v_2 $

The formula for average velocity is:

$ v_{\text{avg}} = \frac{\text{Total Distance}}{\text{Total Time}} $

Given:

$ v_{\text{avg}} = \frac{50}{7} \, \text{m/s} $

The total distance traveled is:

$ x + \frac{3x}{2} = \frac{5x}{2} $

The time taken to cover each segment is given by:

Time for $ x $: $ t_1 = \frac{x}{v_1} = \frac{x}{5} $

Time for $ \frac{3x}{2} $: $ t_2 = \frac{\frac{3x}{2}}{v_2} = \frac{3x}{2v_2} $

Now, using the formula for average velocity:

$ \frac{50}{7} = \frac{\frac{5x}{2}}{\frac{x}{5} + \frac{3x}{2v_2}} $

Simplifying the equation:

$ \frac{50}{7} = \frac{5/2}{\frac{1}{5} + \frac{3}{2v_2}} $

Cross-multiplying gives:

$ \frac{1}{5} + \frac{3}{2v_2} = \frac{7}{20} $

Solving for $ \frac{3}{2v_2} $:

$ \frac{3}{2v_2} = \frac{7}{20} - \frac{1}{5} = \frac{7-4}{20} = \frac{3}{20} $

Finally, solving for $ v_2 $:

$ \frac{3}{2v_2} = \frac{3}{20} $

$ v_2 = 10 \, \text{m/s} $

Thus, the value of $ v_2 $ is $ 10 \, \text{m/s} $.

Here, we will use the concept of relative motion.

Let initial (at $t = 0$) position of both cars is x = 0 (As they cross each other at t = 0 for Ist time)

given : For P

${a_P} \propto t$

For Q,

${a_Q} = a'$ (Let) (constant)

$ \Rightarrow {a_{{p^{(t)}}}} = kt$ where, k = constant

$ \Rightarrow {{d{v_p}(t)} \over {dt}} = kt$ (as $a = {{dv} \over {dt}}$)

$ \Rightarrow \int_0^{{v_p}(t)} {d{v_p}(t) = \int_0^t {ktdt} } $ [let p and Q both starts from rest]

$ \Rightarrow {v_p}(t) = {{k{t^2}} \over 2}$

$ \Rightarrow {{d{x_p}(t)} \over {dt}} = {{k{t^2}} \over 2}$ [As $v = {{dx} \over {dt}}$]

$ \Rightarrow \int_0^{{x_p}(t)} {d{x_{p(t)}} = \int_0^t {{{k{t^2}} \over 2}dt} } $

$ \Rightarrow {x_p}(t) = {{k{t^3}} \over 6}$ .... (1)

Now, for Q,

${x_Q}(t) = {1 \over 2}{a^1}{t^2}$ .... (2) [using Newton's 2nd equation of motion]

So, relative position of P w.r.t. Q,

${x_{P{Q^{(t)}}}} = {x_p}(t) - {x_Q}(t)$

$ \Rightarrow {x_{P{Q^{(t)}}}} = {{k{t^3}} \over 6} - {1 \over 2}{a^1}{t^2}$ (From (1) and (2))

When both cars cross each other, $X_{PQ}=O$

As $X_{PQ}(t)$ is a cubic polynomial, so it has maximum 3 roots.

Hence, the maximum number of crossing = 3

Given the displacement of the particle $x^2 = 1 + t^2$, we want to find the acceleration, which is the second derivative of displacement with respect to time, $a = \frac{d^2x}{dt^2}$, and we are given that the acceleration at any time $t$ is $x^{-n}$, for us to find the value of $n$.

First, let's find the first derivative of displacement with respect to time, which gives us the velocity. Differentiating $x^2 = 1 + t^2$ with respect to t, we get:

$2x\frac{dx}{dt} = 2t$

This simplifies to:

$\frac{dx}{dt} = \frac{t}{x}$

Now, let's differentiate this velocity to find the acceleration:

$a = \frac{d^2x}{dt^2} = \frac{d}{dt}\left(\frac{t}{x}\right)$

To differentiate $\frac{t}{x}$ with respect to $t$, we'll use the quotient rule:

$\frac{d}{dt}\left(\frac{t}{x}\right) = \frac{x\cdot 1 - t\cdot \frac{dx}{dt}}{x^2}$

Substitute $\frac{dx}{dt} = \frac{t}{x}$ into the equation:

$a = \frac{x(1) - t(\frac{t}{x})}{x^2} = \frac{x - \frac{t^2}{x}}{x^2} = \frac{x^2 - t^2}{x^3}$

Recalling that the displacement equation given was $x^2 = 1 + t^2$, substitute this into our expression for acceleration:

$a = \frac{1 + t^2 - t^2}{x^3} = \frac{1}{x^3}$

Thus, the acceleration of the particle at any time $t$ is $x^{-3}$, which means our value for $n$ is $3$.

Given that a body is moving on a frictionless plane and starts from rest, the motion can be assumed to be uniformly accelerated motion. The formula for the distance covered in uniformly accelerated motion from rest is given by $s = ut + \frac{1}{2}at^2$, where:

• $s$ is the distance covered,


• $u$ is the initial velocity (which is 0 since the body starts from rest),


• $a$ is the acceleration, and


• $t$ is the time.

Since the body starts from rest ($u=0$), the formula simplifies to $s = \frac{1}{2}at^2$.

The distance moved between $t = n - 1$ and $t = n$, denoted as $S_n$, can be found by calculating the distance covered by the end of time $n$ and subtracting the distance covered by the end of time $n-1$. Let's denote the total distance covered by time $n$ as $S(n)$, which according to the formula is $S(n) = \frac{1}{2}a n^2$. Thus, $S_n = S(n) - S(n-1)$.

Therefore, we have:

$S_n = \frac{1}{2}an^2 - \frac{1}{2}a(n-1)^2$

Simplifying this, we get:

$S_n = \frac{1}{2}a(n^2 - (n^2 - 2n + 1))$

This simplifies to:

$S_n = \frac{1}{2}a(2n - 1)$

Similarly,

$S_{n-1} = \frac{1}{2}a((n-1)^2 - (n-2)^2)$

Simplifying:

$S_{n-1} = \frac{1}{2}a((n-1)^2 - (n^2 - 4n + 4))$

Which further simplifies to:

$S_{n-1} = \frac{1}{2}a(2n - 3)$

Hence, the ratio $\frac{S_{n-1}}{S_n}$ can be calculated:

$\frac{S_{n-1}}{S_n} = \frac{\frac{1}{2}a(2n - 3)}{\frac{1}{2}a(2n - 1)} = \frac{2n - 3}{2n - 1}$

For $n = 10$,

$\frac{S_{n-1}}{S_n} = \frac{2(10) - 3}{2(10) - 1} = \frac{20 - 3}{20 - 1} = \frac{17}{19}$

Given that the ratio is represented as $\left(1 - \frac{2}{x}\right)$, we have:

$\frac{17}{19} = 1 - \frac{2}{x}$

Solving this equation for $x$ gives:

$1 - \frac{17}{19} = \frac{2}{x}$

$\frac{2}{19} = \frac{2}{x}$

Thus:

$x = 19$

A bus traveling along a straight highway at a speed of 72 km/h comes to a stop within 4 seconds after the brakes are applied. To find the distance the bus travels during this time (assuming uniform deceleration), follow these steps:

First, convert the initial speed from km/h to m/s:

$\begin{aligned} u &= 72 \times \frac{5}{18} = 20 \, \text{m/s} \end{aligned}$

Given:

Initial speed ($ u $) = 20 m/s

Final speed ($ v $) = 0 m/s

Time ($ t $) = 4 s

Using the first equation of motion:

$\begin{aligned} v &= u + at \\ 0 &= 20 + 4a \\ a &= -5 \, \text{m/s}^2 \end{aligned}$

Now, use the second equation of motion to find the distance traveled ($ S $):

$\begin{aligned} S &= ut + \frac{1}{2}at^2 \\ S &= 20 \times 4 + \frac{1}{2} \times (-5) \times 4^2 \\ S &= 80 - 40 \\ S &= 40 \, \text{m} \end{aligned}$

Therefore, the distance traveled by the bus during the time the brakes are applied is 40 meters.

To find the acceleration of the particle, we first need to differentiate the velocity function with respect to time. The velocity function given is

$ v = 4\sqrt{x} $

However, this function gives the velocity as a function of position $x$, not as a function of time $t$. Since acceleration is the rate of change of velocity with respect to time, we'll need to use the chain rule to differentiate $v$ with respect to $t$.

The chain rule in this context can be stated as follows:

$ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} $

Now, because $\frac{dx}{dt}$ is the velocity $v$ itself and $\frac{dv}{dx}$ is the derivative of the velocity with respect to $x$, we first find $\frac{dv}{dx}$:

$ v = 4\sqrt{x} = 4x^{\frac{1}{2}} $

Differentiating with respect to $x$, we get:

$ \frac{dv}{dx} = 4 \cdot \frac{1}{2} x^{-\frac{1}{2}} = 2x^{-\frac{1}{2}} = \frac{2}{\sqrt{x}} $

Now, because $v = 4\sqrt{x}$, we can rewrite $\sqrt{x}$ as $\frac{v}{4}$. Using this to replace $\sqrt{x}$ in our expression for $\frac{dv}{dx}$, we get:

$ \frac{dv}{dx} = \frac{2}{\sqrt{x}} = \frac{2}{\frac{v}{4}} = \frac{8}{v} $

Now, using the chain rule:

$ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{8}{v} \cdot v $

Simplifying this, the velocity terms cancel out, leaving us with:

$ a = 8 \text{ ms}^{-2} $

Thus, the acceleration of the particle is $8 \text{ ms}^{-2}$.

• Position (x): The particle's location on the x-axis at a given time.


• Velocity (v): The rate of change of position with respect to time

( $v = \frac{dx}{dt}$ ).

• Acceleration (a): The rate of change of velocity with respect to time

( $a = \frac{dv}{dt}$ ).

The particle's position is given by:

$x(t) = -3t^3 + 18t^2 + 16t$

We need to find the velocity when the acceleration is zero.

1. Find the Velocity (v) and Acceleration (a) Functions:

• Velocity:

$v(t) = \frac{dx}{dt} = -9t^2 + 36t + 16$

• Acceleration:

$a(t) = \frac{dv}{dt} = -18t + 36$

1. Find the Time (t) When Acceleration is Zero:

$a(t) = 0$

$-18t + 36 = 0$

$t = 2 \text{ seconds}$

1. Calculate the Velocity at t = 2 seconds:

$v(2) = -9(2)^2 + 36(2) + 16$

$v(2) = 52 \text{ m/s}$

The velocity of the particle when its acceleration becomes zero is 52 m/s.

The displacement and the increase in the velocity of a moving particle from time $t$ to $(t + 1) \mathrm{s}$ are $125 \mathrm{~m}$ and $50 \mathrm{~m} / \mathrm{s}$, respectively. The distance traveled by the particle in $(\mathrm{t} + 2)^{\mathrm{th}} \mathrm{s}$ is calculated as follows:

Given that the acceleration is constant, we start with:

$ v = u + at $

When the velocity has increased by $50 \mathrm{~m}/\mathrm{s}$, the equation becomes:

$ u + 50 = u + a \quad \Rightarrow \quad a = 50 \mathrm{~m}/\mathrm{s}^2 $

Next, we consider the displacement:

$ 125 = u t + \frac{1}{2} a t^2 $

Since this is given over a unit time interval (from $t$ to $(t + 1)$), we use:

$ 125 = u + \frac{a}{2} $

Substituting $a = 50 \mathrm{~m}/\mathrm{s}^2$:

$ 125 = u + \frac{50}{2} \quad \Rightarrow \quad 125 = u + 25 \quad \Rightarrow \quad u = 100 \mathrm{~m}/\mathrm{s} $

To find the distance traveled by the particle in $(t + 2)^\text{th}$ second, we use:

$ S_n = u + \frac{a}{2} [2n - 1] $

For $n = t + 2$ (i.e., the (t+2)th second):

$ S_{(t+2)} = u + \frac{a}{2} [2(t+2) - 1] $

With $u = 100$ and $a = 50$:

$ S_{(t+2)} = 100 + \frac{50}{2} [2(t+2) - 1] = 100 + 25 \times [2(t+2) - 1] = 100 + 25 \times (2t + 4 - 1) = 100 + 25 \times (2t + 3) = 100 + 25 \times 5 = 100 + 125 = 225 \mathrm{~m} $

To find the distance of the upper point A from the starting point, we need to first understand the motion of a freely falling body under the influence of gravity. The body falling under gravity is an example of uniformly accelerated motion with the acceleration equal to the acceleration due to gravity, which is given as $ g = 10 \textrm{m/s}^2 $.

We will consider point A where the body was at time $ t_1 $ and point B where the body was at time $ t_2 = t_1 + 2 \textrm{s} $. The displacement in these 2 seconds is given to be 80 meters.

For an object under constant acceleration, the displacement $ s $ can be found using the equation:

$ s = ut + \frac{1}{2}at^2 $

where

$ u $ is the initial velocity,

$ t $ is the time,

$ a $ is the acceleration,

$ s $ is the displacement.

Since the body falls under gravity, its initial velocity at the starting point is 0 ($ u = 0 $), thus the equation simplifies to:

$ s = \frac{1}{2}gt^2 $

Let's denote the distance of point A from the starting point as $ s_A $ and the distance of point B as $ s_B $. We know the body covers $ 80 $ meters in $ 2 $ seconds from A to B, so we can write:

$ s_B - s_A = 80 \textrm{m} $

We can calculate the distance covered till point B (in time $ t_2 = t_1 + 2 $) as:

$ s_B = \frac{1}{2}g(t_1 + 2)^2 $

And the distance covered till point A (in time $ t_1 $) as:

$ s_A = \frac{1}{2}gt_1^2 $

Now, substituting $ s_B $ and $ s_A $ in our difference equation:

$ \frac{1}{2}g(t_1 + 2)^2 - \frac{1}{2}gt_1^2 = 80 $

$ g \left( \frac{1}{2}(t_1 + 2)^2 - \frac{1}{2}t_1^2 \right) = 80 $

$ 5((t_1 + 2)^2 - t_1^2) = 80 $

$ (t_1 + 2)^2 - t_1^2 = 16 $

$ t_1^2 + 4t_1 + 4 - t_1^2 = 16 $

$ 4t_1 + 4 = 16 $

$ 4t_1 = 16 - 4 $

$ 4t_1 = 12 $

$ t_1 = 3 \textrm{s} $

Therefore, time $ t_1 $ at point A is $ 3 $ seconds. To find $ s_A $, we can use the above simplified motion equation:

$ s_A = \frac{1}{2}gt_1^2 $

$ s_A = \frac{1}{2} \times 10 \times 3^2 $

$ s_A = 5 \times 9 $

$ s_A = 45 \textrm{m} $

So, the distance of the upper point A from the starting point is $ \mathbf{45 \textrm{m}} $.

Alternate Method :

$ \Rightarrow {t_1} = {{{S \over 2}} \over 5}$ ........ (1)

Also, ${S \over 2} = {{10{t_2}} \over 2} + {{15{t_2}} \over 2}$

$ \Rightarrow {t_2} = {S \over {25}}$ ....... (2)

$\Rightarrow$ Mean speed $ = {S \over {{t_1} + {t_2}}}$

$ = {S \over {{S \over {10}} + {S \over {25}}}} = {{250} \over {35}}$ m/s $ = {{50} \over 7}$ m/s

v = 19.6 m/s

t = 6s

Time taken in upward motion above tower = 2s

$\Rightarrow$ Time taken from top most point to ground = 4s

$ \Rightarrow \sqrt {{{2h} \over g}} = 4$

$h = {{16 \times 9.8} \over 2} = 8 \times 9.8$

$ \Rightarrow k = 8 \times 9.8 \times 5 = 392$

${{dv} \over {dx}} = 5$ ms^$-$1/m

Acceleration of particle

when $v = 20$ m/s

$a = v{{dv} \over {dx}} = 20(5)$ m/s² = 100 m/s²

${F_R}\,d = {1 \over 2}m{v^2}$

${{{d_2}} \over {{d_1}}} = {\left( {{{{v_2}} \over {{v_1}}}} \right)^2} = {\left( {{1 \over 3}} \right)^2}$

${d_2} = {d_1} \times {1 \over 9} = 3m$

${v_{mean}} = {{3{v_1}{v_2}{v_3}} \over {{v_1}{v_2} + {v_2}{v_3} + {v_3}{v_1}}}$

$ = {{3 \times 11 \times 22 \times 33} \over {11 \times 22 + 22 \times 33 + 33 \times 11}}$

$ = 18$ m/sec

At t = 2 s, v₁ = 50 $-$ 2 $\times$ 10 = 30 m/s

v₂ = v₂

$\therefore$ a_rel = g $-$ g = 0

$S = {{{u^2} - {v^2}} \over {2g}} = {{{{50}^2} - {{30}^2}} \over {2 \times 10}} = {{1600} \over {20}} = 80$ m

$\therefore$ v_rel = 50 $-$ 30 = 20 m/s

$\therefore$ $\Delta t = {{80} \over {20}} = 4\,s$

$\therefore$ required time t = 2 + 4 = 6 s

Let at h height its velocity becomes 10 m/s (as given in question)

$ \Rightarrow {v^2} - {u^2} = 2( - 10) \times ( - h)$

$ \Rightarrow {v^2} = 20\,h$

$ \Rightarrow 100 = 20\,h$

$ \Rightarrow h = 5\,m$

$h' = 10 - 5 = 5\,m$

Based on the situation

$h = - u{t_1} + {1 \over 2}gt_1^2$ $\to$ throwing up ....... (i)

$h = u{t_2} + {1 \over 2}gt_2^2$ $\to$ throwing up ....... (ii)

$h = {1 \over 2}g{t^2}$ $\to$ dropping .......... (iii)

and $0 = u({t_1} - {t_2}) - {1 \over 2}g{({t_1} - {t_2})^2}$ ....... (iv)

solving above equations

$t = \sqrt {{t_1}{t_2}} $

$ \Rightarrow t = \sqrt {6 \times 1.5} = 3\,s$